Since B0 = 1 and [itex] \frac { { \partial }^{ N }f }{ \partial { B }^{ N } } [/itex] can be treated as scalar derivatives of a normal function (a number) and BN was shown to be Hermitian, we're left with a scalar multiplied by a Hermitian, which is itself Hermitian.

As for ##B^*##, it is the Hermitian conjugate. In this case, I probably should be using the dagger notation ##{ B }^{ \dagger }##.

Sure. So Hermitian means ##<f|Bg>=<Bf|g>## for any f and g. That makes it 'almost' obvious that ##<f|B^ng>=<B^nf|g>##. If it's not sufficiently obvious, it's easy to formalize the proof with induction.